Compare the object distance with f first

Compare $s_o$ with $f$ before predicting anything.

1. $s_o > f$: the rays converge past the lens and cross. Real, inverted image on the far side. Enlarged if the object is between $f$ and $2f$, reduced if beyond $2f$.
2. $s_o < f$: the rays are still diverging when they leave, so they never cross. Virtual, upright, enlarged image on the object's side.

Take $s_o = 4$ cm with $f = 10$ cm. The equation gives $s_i = -6.7$ cm: an upright virtual image $6.7$ cm out on the object's side, about $1.7$ times as tall. Nothing is wrong with the problem, and the negative sign is the equation reporting a virtual image, exactly as it does for mirrors.

"Converging means real" is a rule with a condition attached, and dropping the condition is what turns a magnifying glass into an error message. Inside the focal point is the only way a single converging optic enlarges while staying upright.

A diverging lens has no such split: with $f < 0$ every object distance gives an upright, reduced, virtual image, which is the same fixed outcome a convex mirror has.

$$\frac{1}{s_i} + \frac{1}{s_o} = \frac{1}{f}.$$

Two failures, both mechanical:

1. Subtracting the distances directly: writing $s_i = f - s_o$. With $f = 10$ cm and $s_o = 15$ cm that gives $-5$ cm and a virtual image, when the true answer is $+30$ cm and real.
2. Forgetting the final inversion: reaching $1/s_i = 1/30$ per cm and reporting $0.033$ cm. An image a fraction of a millimetre from the lens is the tell.

Do it correctly: $1/s_i = 1/10 - 1/15 = 1/30$, so $s_i = 30$ cm, real and on the far side.

Then test the behaviour, not just the number. Sliding the object toward the focal point must drive the image further away, heading to infinity as $s_o \to f$. An answer that moves the image closer as the object approaches $f$ has an algebra error in it whatever the arithmetic says.

The three principal rays for a thin lens, all leaving the same point, the tip of the object:

1. In parallel to the axis, out through the far focal point.
2. Straight through the centre of the lens, undeviated.
3. In through the near focal point, out parallel to the axis.

Each bends exactly once, at the single vertical line through the lens centre. Not at the two glass faces, which is a real refraction the thin-lens approximation deliberately collapses into one, and not at a focal point, which is somewhere a ray passes through rather than a place it turns.

Mark the image tip where any two of them meet, and put the image base directly below it on the axis.

Rays drawn from the base of the object run along the axis, cross everywhere, and locate nothing, which is why starting at the tip is the first instruction rather than a detail. Rays hinged at a focal point put the image on the wrong side entirely.

Once you have $s_i$, the rest follows without further decisions. The CED writes the size ratio with bars throughout:

$$|M| = \left|\frac{h_i}{h_o}\right| = \left|\frac{s_i}{s_o}\right|,$$

and this course carries orientation in the sign of $M = h_i/h_o$, negative for inverted and positive for upright.

1. $s_i > 0$: real, on the far side, inverted.
2. $s_i < 0$: virtual, on the object's side, upright.
3. $|M| > 1$: enlarged. $|M| < 1$: reduced.

For a single lens those readings always agree with each other: real with inverted, virtual with upright. When they disagree, one of the two was computed wrongly, so it is worth checking them against each other before writing anything down.